Show that a polynomial over $\mathbb{Z}_{2}$ is irreducible Given the polynomial: $p(x)=x^4+x^3+x^2+x+1$ over $\mathbb{Z}_{2}$, to show that it is irreducable, is it enough to show that $p(0)=p(1)=1$?
 A: No: A solution $r \in \mathbb{Z}_2$ to $p(x) = 0$ corresponds to a factor $x - r$, so showing that $p(0) = p(1) = 1$, i.e., that there is no root, simply shows that $p$ does not have any linear factors. Since $p$ is quartic, however, a priori it might admit a factorization without any linear factors, that is, into a product of two (irreducible) quadratic polynomials.
Hint On the other hand, we can quickly identify all of the irreducible quadratic polynomials: The reducible quadratic polynomials are the products of two linear factors, namely, $x^2$, $x (x + 1) = x^2 + x$, and $(x + 1)^2 = x^2 + 1$, and so the only irreducible quadratic polynomial is $x^2 + x + 1$. 

So, the only quartic which factors into a product of two irreducible quadratics is $$(x^2 + x + 1)^2 = x^4 + x^2 + 1.$$ This is not our polynomial, which hence admits no such factorization, and which thus is irreducible.

A: On factorising into quadratics, just try by hand. The coefficient of $x^2$ has to be $1$ in both factors to give $x^4$ and the constant in each factor has to be $1$ too to give a constant term $1$ in the product.
So $$(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+abx^2+(a+b)x+1=x^4+x^3+x^2+x+1$$
We have $a+b=1$ so $a$ and $b$ have different parities, and one is $0$. But then $ab=0$ and the coefficient of $x^2$ is zero. So there is no possible factorisation.
A: As you already checked, it has not linear factors so it can only have quadratic ones...but the only irreducible quadratic over $\;\Bbb F_2\;$ is $\;x^2+x+1\;$ , and you can check it does not divide your quartic since:
$$x^4+x^3+x^2+x+1=(x^2+x+1)^2+x(x+1)$$
A: $p$ has got no roots in $\mathbb{Z}_2$ because $p(0)=p(1)=0$. If it were reducible it could only factorise into two quadratics. And because its highest and lowest degree coefficient is $1$ their highest and lowest degree coefficient should be $1$.
Assume $p(x)=(x^2+ax+1)(x^2+bx+1)$. Identifying the coefficients this leads to
$$\begin{cases} a+b=1\\ab=1\end{cases}$$
And this is impossible in $\mathbb{Z}_2$
